This is a follow-up to this question. So that you don't have to flick back and forwards I'll briefly summarize:

My original question was on how to prove that a polynomial $g(x)$ obtained from $f(t,x)$ by specializing $t$ to some rational value has Galois group over $\mathbb Q$ a subgroup of that of $f$ over $\mathbb{Q}(t)$. Keith Conrad wrote out a nice proof using the factorization of prime ideals in extensions of Dedekind domains.

I have been trying to extend this result to the multivariate case, to show that eg. given some polynomial $f(s,t,x)$ in $\mathbb{Q}[s,t,x]$, a rational specialization of $s$ will result in a polynomial with Galois group over $\mathbb{Q}(t)$ which is a subgroup of that of $f$ over $\mathbb{Q}(s,t)$.

Unfortunately it is not as simple as replacing $\mathbb{Q}[t]$ with $\mathbb{Q}[s,t]$ in Keith's proof and considering the splitting of the prime ideal $(s-s_0)$ (where $s_0$ is the rational value being substituted in for $s$). $\mathbb{Q}[s,t]$ is not a Dedekind domain, and so the proof would fall through in various places, not least because $(s-s_0)$ is not a maximal ideal.

Does anyone know if this result still holds for Galois groups over function fields in more than one variable? And more generally, does unique factorization of prime ideals hold for any non-Dedekind domains?

A domain other than a field in which nonzero proper ideals factor uniquely into a product of prime ideals is a Dedekind domain. In fact, a domain other than a field in which every nonzero ideal is a product of prime ideals is a Dedekind domain (so existence of prime ideal factorizations all the time implies uniqueness). The correct extension of Dedekind domains as far unique factn. is concerned is Krull domains, e.g., if A is Krull then A[x] is also Krull.
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KConradSep 8 '10 at 17:16

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Dear Adam: For consistency with the notation in the earlier version, let's speak in terms of specialization of $t$ rather than $s$. Set $L = \mathbf{Q}(s)$ in the SGA1 argument that I gave in a comment to your question. That involves specializing $t$ into $\mathbf{Q}(s)$ rather than into $\mathbf{Q}$, but if you look at the proof you'll see that it provides an abundance of rational specializations of $t$ as well (in fact, all but finitely many). The argument is quite robust.
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BCnrdSep 8 '10 at 17:17

So now I have to learn algebraic geometry as well? This problem is going to have taught me a lot of mathematics.... Thankyou Conrads!
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Adam Sep 8 '10 at 17:36

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Adam, to paraphrase Hilbert, to do algebraic number theory without any algebraic geometry is like boxing without fists. (Strictly speaking, the content of my earlier comment can be done by pure algebra without any algebraic geometry. But that would be kind of silly, since it removes all of the insight which leads one to the argument in the first place.)
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BCnrdSep 8 '10 at 18:15

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Adam, concerning your pacifist tendencies, please go to the library (or some website of illegal scans, if the concept of library is too archaic) and compare two books on algebraic curves: the one by Fulton and the one by Chevalley. I suspect that a pacifist would become violent in rather different senses after looking at each of those books. :)
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BCnrdSep 8 '10 at 22:05